# Question db883

Aug 10, 2017

83.8%

#### Explanation:

Start by writing the balanced chemical equation that describes this reaction

${\text{Fe"_ 2"O"_ (3(s)) + 3"CO"_ ((g)) -> 2"Fe"_ ((s)) + 3"CO}}_{2 \left(g\right)}$

Now, notice that for every $1$ mole of iron(III) oxide that takes part in the reaction, the reaction can theoretically produce $2$ moles of iron.

This represents the reaction's theoretical yield, i.e. what you get for a reaction that has an 100% yield.

In your case, you know that $\text{150.0 g}$ of iron(III) oxide react with excess carbon monoxide, which means that all the mass of iron(III) oxide take part in the reaction.

Use the molar mass of iron(III) oxide to convert the sample to moles

150.0 color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_3)/(159.69color(red)(cancel(color(black)("g")))) = "0.9393 moles Fe"_2"O"_3

Now, this much iron(III) oxide would theoretically produce

0.9393 color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * overbrace("2 moles Fe"/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))))^(color(blue)("from the balanced chemical equation")) = "1.879 moles Fe"

This would be equivalent to

1.897 color(red)(cancel(color(black)("moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = "104.93 g"

So, the reaction should produce $\text{104.93 g}$ of iron at 100%. However, you know that it only produces $\text{87.9 g}$ of iron--this represents the actual yield of the reaction.

You can thus say that the reaction has a percent yield equal to

87.9 color(red)(cancel(color(black)("g Fe"))) * "100% yield"/(104.93 color(red)(cancel(color(black)("g Fe")))) = color(darkgreen)(ul(color(black)(83.8%)))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of iron produced by the reaction.